Math, asked by channaveeresh, 10 months ago

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Answered by johangeo71
1

Answer:

Answer:

The volume of water left in the tube is 616 cm³.

Step-by-step explanation:

SOLUTION :  

Given :  

Radius of the cylindrical tub (r) = 5 cm

Height of the Cylindrical tub ( H) = 9.8 cm

Height of the cone outside the hemisphere (h) = 5 cm

Radius of the hemisphere = 5 cm

Volume of the cylindrical tub , V1  = πr²H

V1 = π( 5)² ×  9.8 = 22/7 × 25 × 9.8 = 22 × 25 × 1.4

V1 = 770 cm³

Volume of the hemisphere , V2 = 2/3πr³

V2 = ⅔ × 22/7 × 3.5³

V2 = 89.79 cm³

Volume of the cone  , V3 = ⅓πr²h

V3 = ⅓ × 22/7 ×3.5² × 5 = (22 × 3.5 × 3.5 ×5)/ (3 × 7) = (22 × 0.5 × 3.5 ×5)/ 3 = 192.5/3 = 64.14 cm³

V3 = 64.14 cm³

Total volume of the solid immersed in the tub = Volume of the cone + Volume of the hemisphere  = V2 + V3

V = 89.79 + 64.14 =  154 cm³

Hence, the total volume of the solid immersed in the tub = 154 cm³

Volume of the water left in the tub = volume of the cylinder - Total volume of the solid immersed in the tub

= 770 - 154 = 616 cm³

Hence, the volume of water left in the tube is 616 cm³.

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